I have a unit test that needs to work with XML file located in src/test/resources/abc.xml. What is the easiest way just to get the content of the file into String?
Finally I found a neat solution, thanks to Apache Commons:
package com.example;
import org.apache.commons.io.IOUtils;
public class FooTest {
#Test
public void shouldWork() throws Exception {
String xml = IOUtils.toString(
this.getClass().getResourceAsStream("abc.xml"),
"UTF-8"
);
}
}
Works perfectly. File src/test/resources/com/example/abc.xml is loaded (I'm using Maven).
If you replace "abc.xml" with, say, "/foo/test.xml", this resource will be loaded: src/test/resources/foo/test.xml
You can also use Cactoos:
package com.example;
import org.cactoos.io.ResourceOf;
import org.cactoos.io.TextOf;
public class FooTest {
#Test
public void shouldWork() throws Exception {
String xml = new TextOf(
new ResourceOf("/com/example/abc.xml") // absolute path always!
).asString();
}
}
Right to the point :
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
Assume UTF8 encoding in file - if not, just leave out the "UTF8" argument & will use the
default charset for the underlying operating system in each case.
Quick way in JSE 6 - Simple & no 3rd party library!
import java.io.File;
public class FooTest {
#Test public void readXMLToString() throws Exception {
java.net.URL url = MyClass.class.getResource("test/resources/abc.xml");
//Z means: "The end of the input but for the final terminator, if any"
String xml = new java.util.Scanner(new File(url.toURI()),"UTF8").useDelimiter("\\Z").next();
}
}
Quick way in JSE 7
public class FooTest {
#Test public void readXMLToString() throws Exception {
java.net.URL url = MyClass.class.getResource("test/resources/abc.xml");
java.nio.file.Path resPath = java.nio.file.Paths.get(url.toURI());
String xml = new String(java.nio.file.Files.readAllBytes(resPath), "UTF8");
}
Quick way since Java 9
new String(getClass().getClassLoader().getResourceAsStream(resourceName).readAllBytes());
Neither intended for enormous files though.
First make sure that abc.xml is being copied to your output directory. Then you should use getResourceAsStream():
InputStream inputStream =
Thread.currentThread().getContextClassLoader().getResourceAsStream("test/resources/abc.xml");
Once you have the InputStream, you just need to convert it into a string. This resource spells it out: http://www.kodejava.org/examples/266.html. However, I'll excerpt the relevent code:
public String convertStreamToString(InputStream is) throws IOException {
if (is != null) {
Writer writer = new StringWriter();
char[] buffer = new char[1024];
try {
Reader reader = new BufferedReader(
new InputStreamReader(is, "UTF-8"));
int n;
while ((n = reader.read(buffer)) != -1) {
writer.write(buffer, 0, n);
}
} finally {
is.close();
}
return writer.toString();
} else {
return "";
}
}
With the use of Google Guava:
import com.google.common.base.Charsets;
import com.google.common.io.Resources;
public String readResource(final String fileName, Charset charset) throws Exception {
try {
return Resources.toString(Resources.getResource(fileName), charset);
} catch (IOException e) {
throw new IllegalArgumentException(e);
}
}
Example:
String fixture = this.readResource("filename.txt", Charsets.UTF_8)
You can try doing:
String myResource = IOUtils.toString(this.getClass().getResourceAsStream("yourfile.xml")).replace("\n","");
Here's what i used to get the text files with text. I used commons' IOUtils and guava's Resources.
public static String getString(String path) throws IOException {
try (InputStream stream = Resources.getResource(path).openStream()) {
return IOUtils.toString(stream);
}
}
You can use a Junit Rule to create this temporary folder for your test:
#Rule public TemporaryFolder temporaryFolder = new TemporaryFolder();
File file = temporaryFolder.newFile(".src/test/resources/abc.xml");
OK, for JAVA 8, after a lot of debugging
I found that there's a difference between
URL tenantPathURI = getClass().getResource("/test_directory/test_file.zip");
and
URL tenantPathURI = getClass().getResource("test_directory/test_file.zip");
Yes, the / at the beginning of the path without it I was getting null!
and the test_directory is under the test directory.
Using Commons.IO, this method works from EITHER a instance method or a static method:
public static String loadTestFile(String fileName) {
File file = FileUtils.getFile("src", "test", "resources", fileName);
try {
return FileUtils.readFileToString(file, StandardCharsets.UTF_8);
} catch (IOException e) {
log.error("Error loading test file: " + fileName, e);
return StringUtils.EMPTY;
}
}
Related
I am trying to create a simple java program which reads and extracts the content from the file(s) inside zip file. Zip file contains 3 files (txt, pdf, docx). I need to read the contents of all these files and I am using Apache Tika for this purpose.
Can somebody help me out here to achieve the functionality. I have tried this so far but no success
Code Snippet
public class SampleZipExtract {
public static void main(String[] args) {
List<String> tempString = new ArrayList<String>();
StringBuffer sbf = new StringBuffer();
File file = new File("C:\\Users\\xxx\\Desktop\\abc.zip");
InputStream input;
try {
input = new FileInputStream(file);
ZipInputStream zip = new ZipInputStream(input);
ZipEntry entry = zip.getNextEntry();
BodyContentHandler textHandler = new BodyContentHandler();
Metadata metadata = new Metadata();
Parser parser = new AutoDetectParser();
while (entry!= null){
if(entry.getName().endsWith(".txt") ||
entry.getName().endsWith(".pdf")||
entry.getName().endsWith(".docx")){
System.out.println("entry=" + entry.getName() + " " + entry.getSize());
parser.parse(input, textHandler, metadata, new ParseContext());
tempString.add(textHandler.toString());
}
}
zip.close();
input.close();
for (String text : tempString) {
System.out.println("Apache Tika - Converted input string : " + text);
sbf.append(text);
System.out.println("Final text from all the three files " + sbf.toString());
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (SAXException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (TikaException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
If you're wondering how to get the file content from each ZipEntry it's actually quite simple. Here's a sample code:
public static void main(String[] args) throws IOException {
ZipFile zipFile = new ZipFile("C:/test.zip");
Enumeration<? extends ZipEntry> entries = zipFile.entries();
while(entries.hasMoreElements()){
ZipEntry entry = entries.nextElement();
InputStream stream = zipFile.getInputStream(entry);
}
}
Once you have the InputStream you can read it however you want.
As of Java 7, the NIO АРI provides a better and more generic way of accessing the contents of ZIP or JAR files. Actually, it is now a unified API which allows you to treat ZIP files exactly like normal files.
In order to extract all of the files contained inside of a ZIP file in this API, you'd do as shown below.
In Java 8
private void extractAll(URI fromZip, Path toDirectory) throws IOException {
FileSystems.newFileSystem(fromZip, Collections.emptyMap())
.getRootDirectories()
.forEach(root -> {
// in a full implementation, you'd have to
// handle directories
Files.walk(root).forEach(path -> Files.copy(path, toDirectory));
});
}
In Java 7
private void extractAll(URI fromZip, Path toDirectory) throws IOException {
FileSystem zipFs = FileSystems.newFileSystem(fromZip, Collections.emptyMap());
for (Path root : zipFs.getRootDirectories()) {
Files.walkFileTree(root, new SimpleFileVisitor<Path>() {
#Override
public FileVisitResult visitFile(Path file, BasicFileAttributes attrs)
throws IOException {
// You can do anything you want with the path here
Files.copy(file, toDirectory);
return FileVisitResult.CONTINUE;
}
#Override
public FileVisitResult preVisitDirectory(Path dir, BasicFileAttributes attrs)
throws IOException {
// In a full implementation, you'd need to create each
// sub-directory of the destination directory before
// copying files into it
return super.preVisitDirectory(dir, attrs);
}
});
}
}
Because of the condition in while, the loop might never break:
while (entry != null) {
// If entry never becomes null here, loop will never break.
}
Instead of the null check there, you can try this:
ZipEntry entry = null;
while ((entry = zip.getNextEntry()) != null) {
// Rest of your code
}
Sample code you can use to let Tika take care of container files for you.
http://wiki.apache.org/tika/RecursiveMetadata
Form what I can tell, the accepted solution will not work for cases where there are nested zip files. Tika, however will take care of such situations as well.
My way of achieving this is by creating ZipInputStream wrapping class that would handle that would provide only the stream of current entry:
The wrapper class:
public class ZippedFileInputStream extends InputStream {
private ZipInputStream is;
public ZippedFileInputStream(ZipInputStream is){
this.is = is;
}
#Override
public int read() throws IOException {
return is.read();
}
#Override
public void close() throws IOException {
is.closeEntry();
}
}
The use of it:
ZipInputStream zipInputStream = new ZipInputStream(new FileInputStream("SomeFile.zip"));
while((entry = zipInputStream.getNextEntry())!= null) {
ZippedFileInputStream archivedFileInputStream = new ZippedFileInputStream(zipInputStream);
//... perform whatever logic you want here with ZippedFileInputStream
// note that this will only close the current entry stream and not the ZipInputStream
archivedFileInputStream.close();
}
zipInputStream.close();
One advantage of this approach: InputStreams are passed as an arguments to methods that process them and those methods have a tendency to immediately close the input stream after they are done with it.
i did mine like this and remember to change url or zip files
jdk 15
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.net.MalformedURLException;
import java.net.URL;
import java.util.Scanner;
import java.util.stream.Stream;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
import java.io.*;
import java.util.*;
import java.nio.file.Paths;
class Main {
public static void main(String[] args) throws MalformedURLException,FileNotFoundException,IOException{
String url,kfile;
Scanner getkw = new Scanner(System.in);
System.out.println(" Please Paste Url ::");
url = getkw.nextLine();
System.out.println("Please enter name of file you want to save as :: ");
kfile = getkw.nextLine();
getkw.close();
Main Dinit = new Main();
System.out.println(Dinit.dloader(url, kfile));
ZipFile Vanilla = new ZipFile(new File("Vanilla.zip"));
Enumeration<? extends ZipEntry> entries = Vanilla.entries();
while(entries.hasMoreElements()){
ZipEntry entry = entries.nextElement();
// String nextr = entries.nextElement();
InputStream stream = Vanilla.getInputStream(entry);
FileInputStream inpure= new FileInputStream("Vanilla.zip");
FileOutputStream outter = new FileOutputStream(new File(entry.toString()));
outter.write(inpure.readAllBytes());
outter.close();
}
}
private String dloader(String kurl, String fname)throws IOException{
String status ="";
try {
URL url = new URL("URL here");
FileOutputStream out = new FileOutputStream(new File("Vanilla.zip")); // Output File
out.write(url.openStream().readAllBytes());
out.close();
} catch (MalformedURLException e) {
status = "Status: MalformedURLException Occured";
}catch (IOException e) {
status = "Status: IOexception Occured";
}finally{
status = "Status: Good";}
String path="\\tkwgter5834\\";
extractor(fname,"tkwgter5834",path);
return status;
}
private String extractor(String fname,String dir,String path){
File folder = new File(dir);
if(!folder.exists()){
folder.mkdir();
}
return "";
}
}
.Net guy working on a java app
I am uploading using the following example as my starting point ( I have this working). This shows using FileContent needing a java.io.File which does not contain the actual file only a pointer to the actual file.
We are uploading from a web site and attempting to insert into the drive, I would prefer to do this using a memory Stream like the .Net example. I cannot see that in looking at the FileContent class. So my questing is: Is there a way to insert a file in Google drive that is in memory and not first on the hard drive?
private static File insertFile(Drive service, String title, String description,
String parentId, String mimeType, String filename) {
// File's metadata.
File body = new File();
body.setTitle(title);
body.setDescription(description);
body.setMimeType(mimeType);
// Set the parent folder.
if (parentId != null && parentId.length() > 0) {
body.setParents(
Arrays.asList(new ParentReference().setId(parentId)));
}
// File's content.
java.io.File fileContent = new java.io.File(filename);
FileContent mediaContent = new FileContent(mimeType, fileContent);
try {
File file = service.files().insert(body, mediaContent).execute();
// Uncomment the following line to print the File ID.
// System.out.println("File ID: " + file.getId());
return file;
} catch (IOException e) {
System.out.println("An error occured: " + e);
return null;
}
}
Override the AbstractInputStream build your own FileContent
package com;
import java.io.IOException;
import java.io.InputStream;
import com.google.api.client.http.AbstractInputStreamContent;
import com.google.api.client.util.Preconditions;
public class FileContent extends AbstractInputStreamContent {
private InputStream inputStream = null;
private long inputLength = 0;
public FileContent(String type, InputStream pInputStream) throws IOException {
super(type);
this.inputStream = Preconditions.checkNotNull(pInputStream);
this.inputLength = this.inputStream.available();
}
public long getLength() throws IOException {
return this.inputLength;
}
public boolean retrySupported() {
return false;
}
#Override
public InputStream getInputStream() throws IOException {
return this.inputStream;
}
}
I need to copy the files(file name contains special character) from one path to another path using URI. But its throws an error. If its successfully copied, if the filename not contains special character. Could you please advise me how to copy the file name with special character using URI from one path to another path. I have copied the code and error below.
Code:-
import java.io.*;
import java.net.URI;
import java.nio.ByteBuffer;
import java.nio.channels.Channels;
import java.nio.channels.ReadableByteChannel;
import java.nio.channels.WritableByteChannel;
public class test {
private static File file = null;
public static void main(String[] args) throws InterruptedException, Exception {
String from = "file:///home/guest/input/3.-^%&.txt";
String to = "file:///home/guest/output/3.-^%&.txt";
InputStream in = null;
OutputStream out = null;
final ReadableByteChannel inputChannel;
final WritableByteChannel outputChannel;
if (from.startsWith("file://")) {
file = new File(new URI(from));
in = new FileInputStream(file);
}
if (from.startsWith("file://")) {
file = new File(new URI(to));
out = new FileOutputStream(file);
}
inputChannel = Channels.newChannel(in);
outputChannel = Channels.newChannel(out);
test.copy(inputChannel, outputChannel);
inputChannel.close();
outputChannel.close();
}
public static void copy(ReadableByteChannel in, WritableByteChannel out) throws IOException {
ByteBuffer buffer = ByteBuffer.allocateDirect(32 * 1024);
while (in.read(buffer) != -1 || buffer.position() > 0) {
buffer.flip();
out.write(buffer);
buffer.compact();
}
}
}
Error:--
Exception in thread "main" java.net.URISyntaxException: Illegal character in path at index 30: file:///home/maria/input/3.-^%&.txt
at java.net.URI$Parser.fail(URI.java:2829)
at java.net.URI$Parser.checkChars(URI.java:3002)
at java.net.URI$Parser.parseHierarchical(URI.java:3086)
at java.net.URI$Parser.parse(URI.java:3034)
at java.net.URI.<init>(URI.java:595)
at com.tnq.fms.test3.main(test3.java:29)
Java Result: 1
Thanks for looking into this...
You can try to use the java.net.uri.
The filenames should be %-escaped. For example, a space in the actual filename becomes a %20 in the URI. The java.net.URI class can do it for you if you use one of the constructors with several arguments:
new URI("file", null, "/home/guest/input/3.-^%&.txt", null);
See HTTP URL Address Encoding in Java.
The following code saves the page's content to a file:
import java.net.*;
import java.io.*;
public class url
{
public static void main(String[] args)
{
try
{
URL PageUrl;
URLConnection GetConn = null;
GetConn = null;
PageUrl = new URL("https://www.google.ru/");
GetConn = PageUrl.openConnection();
GetConn.connect();
InputStreamReader ReadIn = new InputStreamReader(GetConn.getInputStream());
BufferedReader BufData = new BufferedReader(ReadIn);
String htmlFileName = ("C:\\hello.html");
FileWriter FWriter = new FileWriter(htmlFileName);
BufferedWriter BWriter = new BufferedWriter(FWriter);
String UrlData = null;
while ((UrlData = BufData.readLine()) != null)
{
BWriter.write(UrlData);
BWriter.newLine();
}
BWriter.close();
}
catch(IOException io)
{
System.out.println(io);
}
}
}
But I need the file have the same name as the page of the website, for example, it has to somehow get the name of the web page and assign it as the file's name.
You can use URL.getFile() to get the filename. I.e
...
String htmlFileName = PageURL.getFile();
...
Note that different URLs may refer to the same file: http://example.com/test.html#anch1, http://example.com/test.html, http://example.com/test.html?a=b - all three refer to the same test.html file here. In this case you might want to combine getFile(), getRef() and getQuery() somehow.
It is worth to mention some issues in your code:
start variable names with lowerCase instead of UpperCase;
Close resources in finally blocks. Better, if you use Java 7, use try-with-resources.
So I was wondering if it was possible to write all the console output to a separate file outside of Java? I know about the Printwriter and Filewriter method. However, in my experience those would work if I was using them all within one method, but I don't think I can do that with the code I have right now. Below is what I have...
Java Code
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import org.xml.sax.*;
import org.xml.sax.helpers.DefaultHandler;
import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
public class XMLTagParser extends DefaultHandler
{
private int i;
public XMLTagParser()
{
traverse(new File("C:/Documents and Settings/user/workspace/Intern Project/Proposals/Converted Proposals/Extracted Items"));
}
private static final class SaxHandler extends DefaultHandler
{
private StringBuffer buffer;
private String heading;
private boolean inHeading;
public void startElement(String uri, String localName, String qName, Attributes attrs)
{
if ("w:pStyle".equals(qName))
{
String val = attrs.getValue("w:val");
if (val.contains("Heading"))
{
if (isHeading(val))
{
System.out.println(val);
inHeading = true;
}
}
}
if("w:t".equals(qName))
{
if (inHeading == true)
{
buffer = new StringBuffer();
}
}
}
public void characters(char buff[], int offset, int length) throws SAXException
{
String s = new String(buff, offset, length);
if(buffer != null)
{
buffer.append(s);
heading = heading += s;
}
}
public void endElement(String uri, String localName, String qName)
{
buffer = null;
//if the qName is "w:p" and it is in the heading, print out the heading and then reset
if ("w:p".equals(qName) && inHeading == true)
{
System.out.println(heading);
heading = "";
inHeading = false;
}
}
// method to verify whether element is an actual heading
private static boolean isHeading(String heading)
{
String headingNumber = heading.substring(7,8);
String headingName = heading.substring(0,7);
if (headingName.equals("Heading"))
{
if (headingNumber.equals("1")
|| headingNumber.equals("2")
|| headingNumber.equals("3")
|| headingNumber.equals("4")
|| headingNumber.equals("5")
|| headingNumber.equals("6"))
{
return true;
}
}
return false;
}
}
/*private void writeFile(File file)
{
try
{
PrintWriter out = new PrintWriter(new FileWriter(file + "/" + i++));
out.close();
}
catch (IOException e)
{
e.printStackTrace(System.out);
}
}*/
private void traverse(File directory)
{
//Get all files in directory
File[] files = directory.listFiles();
for (File file : files)
{
if (file.getName().equals("document.xml"))
{
try
{
// creates and returns new instance of SAX-implementation:
SAXParserFactory factory = SAXParserFactory.newInstance();
// create SAX-parser...
SAXParser parser = factory.newSAXParser();
// prints out the current working proposal, traversing up the directory structure
System.out.println(file.getParentFile().getParentFile().getName());
// .. define our handler:
SaxHandler handler = new SaxHandler();
// and parse:
parser.parse(file.getAbsolutePath(), handler);
try
{
// instantiates new printwriter which writes out to a file
PrintWriter out = new PrintWriter(new FileWriter(file.getParentFile().getParentFile() + "/" + i++ + ".txt"));
out.close();
}
catch (IOException e)
{
e.printStackTrace(System.out);
}
}
catch (Exception ex)
{
ex.printStackTrace(System.out);
}
}
else if (file.isDirectory())
{
//It's a directory so (recursively) traverse it
traverse(file);
}
}
}
}
So I've instantiated the printwriter in there, but obviously it's no good if I have nothing to write to it. So I'm not really sure how I can get what's printing out the console to be written to that file. Any ideas? Thanks in advance.
If you really want to you can redirect System.out to any PrintStream like this:
PrintStream stream = new PrintStream("filename.txt");
System.setOut(stream);
If you get into rolling your own file logger you'll spend more time dealing with io issues, rolling files, file sizes, ect. You should use log4j instead! It will handle things like this and make your logging more flexible. It's pretty much the standard for java logging.
The System.out is basically an OutputStream; which by default points to the console. Instead, you could just create a new FileOutputStream instance pointing to the file of your choice, and identify this stream by setting it through System.setOut. That will do it for you, throughout the life-cycle of the program/application. Check this link for a complete code.
Instead of using System.out, you could use a FileWriter, write to it and flush it. It is unclear, why you increment i in your code. I guess you want to write everything to just one file.
Also, it looks like you never write to the Writer that you initialize.
You could keep using System.out, and redirect it using ">" to the file when you invoke the application. You can still retain stderr for direct console output.
Or do you mean something else when you write "outside java"?